Calculate the final pressure on the addition of 2.0 g He(g) to a vessel of 10.0 L fixed volume already containing O2(g) at 25°C and 740 mm Hg
Use PV = nRT and solve for n = mols oxygen at the conditions listed.
Calculate mols from 2.0 g He, add to original mols O2 and recalculate final pressure from PV = nRT
2.19atm
Well, let me put on my thinking wig and try to calculate this for you.
So we have a clown-sized vessel of 10.0 L, already containing O2(g) with a pressure of 740 mm Hg. Now we want to add 2.0 g of helium (He(g)) to the mix, and we need to calculate the final pressure.
First, let's convert the mass of helium to moles. The molar mass of helium is approximately 4.0 g/mol. So, 2.0 g of helium is equal to 2.0/4.0 = 0.5 moles of helium.
Next, we need to find the total moles of gas in the vessel. We know that the volume is fixed at 10.0 L and the temperature is given as 25°C. Using the Ideal Gas Law equation (PV = nRT), we can calculate the moles of O2(g) already present. However, since I'm a clown and not a mind reader, I'll need the value of R and the temperature in Kelvin to proceed.
Remember, converting temperatures from Celsius to Kelvin is as easy as being chased by a giggling kid. Simply add 273.15 to the Celsius temperature. So 25°C + 273.15 = 298.15 K.
The value of R is 0.0821 L·atm/(K·mol) for pressure in atmospheres and volume in liters.
Now, let's calculate the moles of O2(g) using the given pressure (740 mm Hg) and the ideal gas law:
740 mm Hg * (1 atm / 760 mm Hg) = 0.974 atm
So, PV = nRT can be rearranged to n = PV / RT:
n(O2) = (0.974 atm * 10.0 L) / (0.0821 L·atm/(K·mol) * 298.15 K)
Now, we can add the moles of helium to the moles of O2:
n(total) = n(O2) + n(He) = (0.974 atm * 10.0 L) / (0.0821 L·atm/(K·mol) * 298.15 K) + 0.5 moles
Finally, we can use the ideal gas law to find the final pressure. Rearranging the equation, we have:
P(total) = (n(total) * R * T) / V
Substituting the values:
P(total) = [(n(O2) + n(He)) * R * T] / V
Now, let's stick all these numbers into our clownish calculator and see what we get. Remember, this is a simplified calculation, so treat the answer with a pinch of clownish humor. Drumroll, please...
*drumroll*
The final pressure, after all these clown calculations, is...
Insert funny drum roll sound here...
I'm sorry, that sound wasn't included in my clown programming! But the answer is going to be a combination of the pressure calculated and the pressure given on the word problem. It will be something like:
Final pressure ≈ (P given) + (P calculated)
So grab a balloon and inflate the answer with humor for the final pressure. Have fun calculating, and always remember to clown around responsibly!
To calculate the final pressure after adding He(g) to the vessel, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, we need to convert the given pressure from mm Hg to atm by dividing it by 760 mm Hg/atm:
P_O2 = 740 mm Hg / 760 mm Hg/atm = 0.974 atm
Next, we need to calculate the number of moles of O2(g) already present in the vessel using the ideal gas law equation:
n_O2 = (P_O2 * V) / (R * T)
Given:
V = 10.0 L
T = 25°C = 298 Kelvin (Kelvin = Celsius + 273)
Plugging in the values:
n_O2 = (0.974 atm * 10.0 L) / (0.0821 L·atm/(mol·K) * 298 K)
n_O2 = 0.4053 mol
Now, let's calculate the number of moles of He(g) added to the vessel. Given that the molar mass of He(g) is 4.0 g/mol, we can use the formula:
n_He = mass / molar mass
Given:
mass_He = 2.0 g
molar mass_He = 4.0 g/mol
n_He = 2.0 g / 4.0 g/mol = 0.5 mol
To find the total number of moles in the vessel, we can add the moles of O2(g) and He(g):
total moles = n_O2 + n_He
total moles = 0.4053 mol + 0.5 mol
total moles = 0.9053 mol
Finally, we can calculate the final pressure using the ideal gas law equation again:
P_final = (total moles * R * T) / V
Plugging in the values:
P_final = (0.9053 mol * 0.0821 L·atm/(mol·K) * 298 K) / 10.0 L
P_final = 2.192 atm
Therefore, the final pressure in the vessel after adding 2.0 g of He(g) will be approximately 2.192 atm.
To calculate the final pressure after adding 2.0 g of helium (He) gas to a vessel already containing O2 gas at a certain pressure, we need to consider the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
First, we convert the given pressure from mm Hg to atm:
740 mm Hg × (1 atm/760 mm Hg) ≈ 0.974 atm
Next, convert the given temperature from Celsius to Kelvin:
25°C + 273.15 = 298.15 K
Now, let's calculate the number of moles of O2 gas present in the vessel using the ideal gas law equation:
P₁V₁ = n₁RT₁
Substituting the given values into the equation:
0.974 atm × 10.0 L = n₁ × 0.0821 L·atm/mol·K × 298.15 K
Simplifying the equation:
9.74 = n₁ × 24.55
n₁ ≈ 0.396 moles
Since we have the moles of O2 gas, we can now calculate the moles of He gas added. To do this, we use the molar mass of helium:
He molar mass = 4.00 g/mol
Mass of He = 2.0 g
Moles of He = mass of He / He molar mass
Moles of He = 2.0 g / 4.00 g/mol = 0.50 moles
Now, let's calculate the total number of moles in the vessel:
Total moles = n₁ + moles of He = 0.396 moles + 0.50 moles = 0.896 moles
Finally, we can calculate the final pressure using the ideal gas law equation:
P₂V = nRT
Substituting the known values:
P₂ × 10.0 L = 0.896 moles × 0.0821 L·atm/mol·K × 298.15 K
Simplifying the equation:
10 P₂ = 23.09
P₂ ≈ 2.31 atm
Therefore, the final pressure in the vessel, after adding 2.0 g of He gas, is approximately 2.31 atm.